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Chapter 9: Problem 33

A uniform bar has two small balls glued to its ends. The bar is \(2.00 \mathrm{~m}\) long and has mass \(4.00 \mathrm{~kg},\) while the balls each have mass \(0.300 \mathrm{~kg}\) and can be treated as point masses. Find the moment of inertia of this combination about an axis (a) perpendicular to the bar through its center; (b) perpendicular to the bar through one of the balls; (c) parallel to the bar through both balls; and (d) parallel to the bar and \(0.500 \mathrm{~m}\) from it.

Short Answer

Expert verified
\( I_{total(a)} = 0.600 \mathrm{~kg \cdot m^2}, I_{total(b)} = 5.200 \mathrm{~kg \cdot m^2}, I_{total(c)} = 0 \mathrm{~kg \cdot m^2}, I_{total(d)} = 1.375 \mathrm{~kg \cdot m^2}\)

Step by step solution

01

Compute inertia for the bar and balls around center of the bar

The bar can be thought of as a point mass at its center when calculating moment of inertia. So for (a), \( I_{bar} = m_{bar}*r_{bar}^2 = 4.00 \mathrm{~kg}*0^2 = 0 \mathrm{~kg \cdot m^2}\). The balls are located at a distance of 1.00 m from the center, so \( I_{balls} = 2*(m_{balls}*r_{balls}^2) = 2*(0.300 \mathrm{~kg}*1.00 \mathrm{~m}^2) = 0.600 \mathrm{~kg \cdot m^2}\). The total moment of inertia \( I_{total(a)} = I_{bar} + I_{balls} = 0.600 \mathrm{~kg \cdot m^2}\).
02

Compute inertia for the bar and balls around one of the balls

The bar is now located 1.00 m away from the axis and one ball is at the axis so it will not contribute to the inertia. So \( I_{bar} = m_{bar}*r_{bar}^2 = 4.00 \mathrm{~kg}*1.00 \mathrm{~m}^2 = 4.00 \mathrm{~kg \cdot m^2}\) and for the other ball, \( I_{balls} = m_{balls}*r_{balls}^2 = 0.300 \mathrm{~kg}*2.00 \mathrm{~m}^2 = 1.200 \mathrm{~kg \cdot m^2}\). Then, \( I_{total(b)} = I_{bar} + I_{balls} = 5.200 \mathrm{~kg \cdot m^2}\).
03

Compute inertia for the bar and balls about an axis through both balls

The bar is now at the axis so its inertia is zero and balls will also not contribute. So, \( I_{total(c)} = 0 \mathrm{~kg \cdot m^2}\).
04

Compute inertia for the bar and balls about an axis parallel to the bar and \(0.500 \mathrm{~m}\) from it

The bar is now located 0.500 m away from the axis and one ball is also at 0.500 m, and the other is 1.500 m away. So \( I_{bar} = m_{bar}*r_{bar}^2 = 4.00 \mathrm{~kg}*0.500 \mathrm{~m}^2 = 1.00 \mathrm{~kg \cdot m^2}\) and for the balls, \( I_{balls} = m_{ball1}*r_{ball1}^2 + m_{ball2}*r_{ball2}^2 = 0.300 \mathrm{~kg}*0.500 \mathrm{~m}^2 + 0.300 \mathrm{~kg}*1.500 \mathrm{~m}^2 = 0.375 \mathrm{~kg \cdot m^2}\). So \( I_{total(d)} = I_{bar} + I_{balls} = 1.375 \mathrm{~kg \cdot m^2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Dynamics
Rotational dynamics is the study of the motion of objects that rotate, and it's a fundamental part of classical mechanics. In rotational dynamics, we often discuss concepts like torque, angular velocity, angular acceleration, and, importantly, moment of inertia. Moment of inertia, symbolized as I, is analogous to mass in linear motion. It measures an object's resistance to changes in its rotational motion based on its mass distribution relative to the axis of rotation.

Understanding moment of inertia is crucial when solving problems related to rotational motion, like the one about the bar and balls. For instance, in solving such a problem, one must consider each component's contribution to the system's overall moment of inertia. It provides a clearer understanding of how mass distribution can drastically change the rotational characteristics of a rigid body.
Physics of Rigid Bodies
The physics of rigid bodies encompasses the study of objects that do not deform under applied forces or, in other words, objects that maintain their shape and size regardless of external influences. In our exercise, the uniform bar with balls glued to its ends can be considered as a system of rigid bodies. Calculations are significantly simplified when we assume bodies are rigid since we can ignore internal structural changes and stresses.

We can easily apply the principles of rotational dynamics to rigid bodies because their shape and mass distribution remain constant, allowing for predictable behavior. This concept allows us to break down complex objects into simpler geometric shapes and point masses to compute properties like center of mass and moment of inertia. It is this simplification that helps in analyzing the moments of inertia in different configurations, without considering the internal structure of the objects.
Parallel Axis Theorem
The parallel axis theorem is an essential tool in the physics of rigid bodies that allows us to find the moment of inertia of an object about any axis, given the moment of inertia about a parallel axis through the object's center of mass and the perpendicular distance between the two axes. The theorem can be expressed mathematically as:
I = Icm + md2
where I is the moment of inertia about the chosen axis, Icm is the moment of inertia about the center of mass axis, m is the mass of the object, and d is the distance between the two parallel axes.

In the given exercise, this theorem is directly applied in part (d) for computing the moment of inertia of the bar and balls about an axis that is parallel to the bar and 0.500 meters away from it. Using this theorem is a significant time-saver, as it prevents the need for complex integration whenever we shift from the center of mass axis to any other parallel axis.

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Most popular questions from this chapter

At \(t=3.00 \mathrm{~s}\) a point on the rim of a \(0.200-\mathrm{m}\) -radius wheel has a tangential speed of \(50.0 \mathrm{~m} / \mathrm{s}\) as the wheel slows down with a tangential acceleration of constant magnitude \(10.0 \mathrm{~m} / \mathrm{s}^{2}\). (a) Calculate the wheel's constant angular acceleration. (b) Calculate the angular velocities at \(t=3.00 \mathrm{~s}\) and \(t=0 .\) (c) Through what angle did the wheel turn between \(t=0\) and \(t=3.00 \mathrm{~s} ?\) (d) At what time will the radial acceleration equal \(g ?\)

A hollow spherical shell has mass \(8.20 \mathrm{~kg}\) and radius \(0.220 \mathrm{~m} .\) It is initially at rest and then rotates about a stationary axis that lies along a diameter with a constant acceleration of \(0.890 \mathrm{rad} / \mathrm{s}^{2}\). What is the kinetic energy of the shell after it has turned through 6.00 rev?

A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev \(/\) min to 200 rev \(/ \min\) in 4.00 s. (a) Find the angular acceleration in rev/s \(^{2}\) and the number of revolutions made by the motor in the 4.00 s interval. (b) How many more seconds are required for the fan to come to rest if the angular acceleration remains constant at the value calculated in part (a)?

\(\mathrm{At} t=0\) a grinding wheel has an angular velocity of \(24.0 \mathrm{rad} / \mathrm{s}\) It has a constant angular acceleration of \(30.0 \mathrm{rad} / \mathrm{s}^{2}\) until a circuit breaker trips at \(t=2.00 \mathrm{~s}\). From then on, it turns through 432 rad as it coasts to a stop at constant angular acceleration. (a) Through what total angle did the wheel turn between \(t=0\) and the time it stopped? (b) At what time did it stop? (c) What was its acceleration as it slowed down?
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